Units Digit

[kevincanspain]

If x is a positive integer, what is the units digit of x^4 -5x^2 + 1 ?

(1) The units digit of x^2 is 4.
(2) The units digit of x^4 is 6.

[Lifetron]

IMO A

[Atekihcan]

x� - 5x² + 1 = (x²)² - 5*(x²) + 1

So, if we know the unit's digit of x² for sure, we can determine the unit's digit of the given expression.

Statement 1 is clearly sufficient.

As for statement 2, unit's digit of (x²)² is 6.
Now, unit's digit of x² can be either 4 or 6. In either case, unit's digit of 5x² is zero. So, the unit's digit of the given expression is = unit's digit of x� + 1 = 6 + 1 = 7
So, statement 2 is also sufficient.

Answer : D

[srcc25anu]

Units digit of x^4 - 5(X^2) + 1

St1: Units digit of x^2 = 4
X^2 can be 4 (x = 2) or x^2 can be 64 (x=8)
If x = 2, x^4 ends in (4*4 = 16) 6 + 1 = 7 and 5*(2^2) ends in 0 (as 5 * even number ends in 0) thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
If x = 8, x^4 ends in (64*64 = 4096) 6 + 1 = 7 and 5*(2^2) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
Hence sufficient

St2: Units digit of x^4 = 6
X can be 2, 4, 6 and 8
If x = 2, x^4 ends in (4*4) 6 + 1 = 7 and 5*(even) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
If x = 4, x^4 ends in (16*16) 6 + 1 = 7 and 5*(even) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
If x = 6, x^4 ends in (36*36) 6 + 1 = 7 and 5*(even) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
If x = 8, x^4 ends in (64*64) 6 + 1 = 7 and 5*(even) ends in 0 thus units digits of x^4 - 5(X^2) + 1 = 7-0 = 7
Hence sufficient

Ans D

[Blue_Skies]

My answer is D.
From 1 only 2 and 8 are the unit digits of the number that qualify from the given statement. Both give the same Unit's digit. Hence answer is D.